Question

$P$ is a point on the line through the point $A$ whose position vector is $\overrightarrow{a}$ and the line is parallel to the vector $\overrightarrow{b}$. If $PA=6$, the position vector of $P$ is

• A. $\overrightarrow{a}+6\overrightarrow{b}$
• B. $\overrightarrow{a}+\frac{6}{\left|\overrightarrow{b}\right|}\overrightarrow{b}$
• C. $\overrightarrow{a}-6\overrightarrow{b}$
• D. $\overrightarrow{b}+\frac{6}{\left|\overrightarrow{a}\right|}\overrightarrow{a}$

Answer 1

Answer: (B)

$\overrightarrow{a}+\frac{6}{\left|\overrightarrow{b}\right|}\overrightarrow{b}$

Line is parallel to vector $\overrightarrow{b}$
The required vector is at a distance of 6 units from $\overrightarrow{a}$
So, the required position vector will be $\overrightarrow{a}+\frac{6}{|\overrightarrow{b}|}\overrightarrow{b}$ since we need a unit vector in the direction of $\overrightarrow{b}$